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Atwood Machine pulley

  1. Jun 13, 2004 #1
    Need closure (if not guidance) on this problem also:

    An Atwood machine consists of two masses, [itex]m_1[/itex] and [itex]m_2[/itex], which are connected by a massless inelastic cord that passes over a pulley. If the puley has radius r and moment of inertia I about its axle, determine the acceleration of the masses [itex]m_1[/itex] and [itex]m_2[/itex].

    The pulley has two forces acting on it which are providing the torque. These two forces are the tensions of the rope. Let [itex]T_1[/itex] and [itex]T_2[/itex] be the tensions caused by [itex]m_1[/itex] and [itex]m_2[/itex] respectively. Using the definition of torque, then
    [tex]
    (T_1 + T_2)r = I\alpha
    [/tex]​
    I'll assume that the angular acceleration of the cord around the pulley is the same as the angular acceleration of the pulley. Since the two masses and the cord have the same acceleration a (magnitude-wise), then I can use [itex]a = \alpha{r}[/itex]. Using F = ma on the two masses, I can solve for the tensions and therefore solve for a in the expression above. I get
    [tex]
    a = \frac{g(m_1 + m_2)}{I/r^2 - m_1 + m_2}
    [/tex]​
     
  2. jcsd
  3. Jun 13, 2004 #2
    You have a sign error in the torque equation. One torque must be negative, the other positive.

    RE: "I'll assume that the angular acceleration of the cord around the pulley is the same as the angular acceleration of the pulley."

    if the cord doesn't slip this must be true, not merely an assumption.
     
  4. Jun 13, 2004 #3

    arildno

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    ehoon, your solution is wrong.
    (However, see comment at the bottom)

    "The pulley has two forces acting on it which are providing the torque. These two forces are the tensions of the rope. Let and be the tensions caused by [tex]m_{1}[/tex] and [tex]m_{2}[/tex] respectively."

    Technically, this is incorrect; it is the friction acting between the cord and the pulley which produces the torque.

    In order to proceed using the force of friction, we must use the condition that the cord is massless.
    (We also assume that the pulley may rotate without friction around its own axle.)
    From Newton's 2.law applied to an arbitrary segment of the cord, it follows that the net sum of forces acting upon the segment must be zero (otherwise, the cord segment would experience infinite acceleration!).

    Let object 1 be on the right-hand side on the pulley, where the cord makes contact with the pulley at [tex]\theta=0[/tex]
    Let object 2 be on the left-hand side on the pulley, where the cord makes contact with the pulley at [tex]\theta=\pi[/tex]

    Regard now a cord segment between [tex]\{\theta,\theta+\bigtriangleup\theta\}[/tex]

    The tangential component of Newton's 2.law for the cord segment reads therefore:
    [tex]T(\theta+\bigtriangleup\theta)-T(\theta)-\bigtriangleup{F}=0[/tex]
    T is the tension in the cord at a given angle; signs are given for the direction in which the local tensile force acts.
    [tex]\bigtriangleup{F}[/tex] is the frictional force acting upon the given segment, and we write it as:
    [tex]\bigtriangleup{F}=f(\theta,t)r\bigtriangleup\theta[/tex]

    f is therefore frictional force per unit length.
    Note that no restrictions are laid upon the functional form of f; it may vary locally with the angle, and also temporally.
    Clearly, the temporal dependence is unnecessary in this case since the driving force, gravity, is time-independent, but if friction about the axle were included, this might well have induced a time-dependence (for example, by using a lubricated shaft as the axle).

    We gain therefore, by going to the limit:
    [tex]\frac{dT}{d\theta}=rf(\theta,t)[/tex]

    Or:
    [tex]T(\pi)=T(0)+r\int_{0}^{\pi}f(\theta,t)d\theta=T(0)+r\overline{f}[/tex]

    Now, using the condition of a massless cord yet again, the tension at object 1 must fulfill:
    [tex]T_{1}=T(0)[/tex]
    And at object 2:
    [tex]T_{2}=T(\pi)[/tex]

    We therefore have Newton's 2.law for the 2 objects:
    [tex]T_{1}-m_{1}g=m_{1}a[/tex]
    [tex]T_{1}+r\overline{f}-m_{2}g=-m_{2}a[/tex]

    Computation of the torque:
    The net torque equals the sum of all frictional torques.
    By Newton's 3.law, we must have that the local force contributing to the torque about the pulley is given by:
    [tex]d\vec{F}=f(\theta,t)rd\theta\vec{i}_{\theta}[/tex]

    Hence, the torque is given by:
    [tex]\vec{\tau}=\int_{0}^{\pi}r\vec{i}_{r}\times\vec{i}_{\theta}rfd\theta[/tex]
    Or, solving for the magnitude:
    [tex]\tau=r^{2}\overline{f}[/tex]

    Hence, for the no-slip case, we have:
    [tex]r^{2}\overline{f}=I\frac{a}{r}[/tex]

    We have now 3 equations and 3 unknowns, and gain:
    [tex]a=\frac{(m_{2}-m_{1})g}{m_{1}+m_{2}+\frac{I}{r^{2}}}[/tex]

    Note that the familiar result with no friction between rope and pulley is obtained by setting I=0

    Comment to your procedure:
    It is seen that if you substitute [tex]T_{2}-T_{1}=r\overline{f}[/tex]
    you will gain the correct expression, as JonDubya said.
    However, I still think my approach is better, since it explicitly includes the friction.
    In addition, if you were to have a slipping rope, it is easier to generalize the technique I have given than the one you used.
     
    Last edited: Jun 13, 2004
  5. Jun 13, 2004 #4

    Gokul43201

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    If arildno scares you...here's something less scary (but less rigorous too)...but I think this is what you want.

    The only mistake you have made is in your first equation. It should instead be:

    [tex]
    (T_2 - T_1)r = I\alpha
    [/tex]​
    (assuming [tex]m_2 > m_1[/tex])

    The free-body diagram tells you why. Here's two alternative explanations :
    1. The two tensions want to rotate the pulley in opposite directions. Hence the difference is the resultant...OR
    2. Use the right hand rule to figure out [tex]T_1 \times r[/tex] and [tex]T_2 \times r[/tex]. They are in opposite directions.

    So, the above equation, along with

    [tex]m_2g - T_2 = m_2a[/tex]

    and

    [tex]T_1 - m_1g = m_1a[/tex]

    Should give the result

    [tex]
    a = \frac{g(m_2 - m_1)}{I/r^2 + m_1 + m_2}
    [/tex]​
     
    Last edited: Jun 13, 2004
  6. Jun 13, 2004 #5

    arildno

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    I don't want to scare anyone, scary people are bad (that's what my Mom told me) :cry:
     
  7. Jun 13, 2004 #6
    I knew I had a sign problem with those tensions.

    Is it static or kinectic friction? I'm going to say static (since the cord isn't slipping).
    I find this a bit sketchy. How can one apply Newton's 2. law if the cord is massless? Do I just plug in 0 for the mass? I understand what you mean though.
    I like your method better too.
     
    Last edited: Jun 13, 2004
  8. Jun 13, 2004 #7
    Double Atwood Machine

    I'm going to mutate my original problem a bit:

    Consider that same Atwood machine as in my original problem, but instead of[itex]m_2[/itex] connected at one end of the cord, I have second Atwood machine there with masses [itex]m_2[/itex] and [itex]m_3[/itex]. What is the acceleration of each mass. Ignore the mass of the pulleys (i.e. consider them as massless).

    I want to solve this problem using Newton's laws only (no energy or Lagrangian/Hamiltonian methods allowed). I drew a free-body diagram for this, but I'm not sure I am analyzing it properly. Intuitively, I know that the second Atwood machine is accelerating, but how can this be if the sum of the forces on the pulley must be zero (since the pulley is massless). There are too many unknowns in this problem and I just don't have enough equations.
     
  9. Jun 13, 2004 #8

    arildno

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    You're right, it's static.
    Yes, you do plug in 0 for the mass.
    Clearly, a real cord is not massless, but if you solve the problem with a non-zero mass (slightly trickier..), you will find that the terms depending on the mass of the cord are negligible to the terms which do not include it.
    That is, the massless cord assumption is a good approximation for small cord masses.
     
  10. Jun 13, 2004 #9

    arildno

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    Note that 0*a=0 for ANY CHOICE OF a!
    (Besides, the Lagrangian is rather neat, you know..:wink:)
     
  11. Jun 14, 2004 #10
    You know what...you're right. So, using F = ma gives me four equations, but then I have six unknowns (the accelerations and tensions). I don't know where to pull out the other equations.

    PS. I bet the Lagrangian is really neat.
     
  12. Jun 14, 2004 #11

    arildno

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    First of all:
    If you choose to use massless pulleys, you cannot have a net, non-zero torque from the force of friction about a given pulley (sinces the moment of inertia is zero).
    Hence, in the massless case, a given rope has equal tension on both sides.
    This gives you four equations and four unknowns.

    If you use pulleys with mass, you have in addition two torque equations.
     
  13. Jun 14, 2004 #12

    Gokul43201

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    It's not hard to see that the tension in the lower rope is just half of that in the higher rope (since the second pulley is massless). So you really have 3 equations in 3 unknowns.
     
  14. Jun 14, 2004 #13
    Wait a moment...I can express any of the three tensions in terms of one tension (one unknown). That leaves me with the acceleration of the three masses (three unknowns). I have three equations of the form F = ma for each mass. Hence, I have four unknowns and three equations. Unless two of the accelerations are equal (which wouldn't make sense), I can't solve this system.
     
  15. Jun 14, 2004 #14

    arildno

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    In the case of massless pulleys:
    As I stated, this imply that the tension in a given cord is equal at every point.
    Since you have 2 cords, that give you 2 unknowns.

    The accelerating pulley is accelerated upwards an equal amount as the 3. mass is accelerated downwards (about the stationary pulley). This follows from the fact that the connecting cord is inextensible; i.e., does not change length.
    Hence, associated by the movement around the stationary pulley, you have one acceleration.

    Go now into the (accelerated) reference frame of the second pulley.
    The two masses on either side must, since their connecting cord is inextensible, have the same magnitude of acceleration, but with opposite sign.

    Hence, you have 4 unknowns in the massless pulley case.

    What are your equations?
    These are, basically, Newton's 2.law applied to the three masses, and Newtons 2. law applied to the accelerating pulley.

    This closes the system.

    Hint:
    It will perhaps be easier to substitute Newton's 2.law applied to the accelerating pulley with Newton's 2.law applying to the accelerating pulley system (that is, the accelerating pulley plus the two masses on either side of it)
    In this case, the tension in the accelerating pulley system will be an internal force; the external forces will be the other tension+forces of gravity.

    I'll post a detailed solution if you still remain confused.
    Good luck!
     
  16. Jun 14, 2004 #15
    I think my problem stems from the fact that this is the first dynamics problem I'm solving where there is an non-inertial reference frame. I don't know how to deal with these kinds of problems since Newton's laws aren't valid in such frames.
    Again, I have no clue how to apply Newton's 2. law properly here. Are you saying that the sum of the forces on the accelerating pulley system equals the tension (excluding the internal ones) minus the weights of the masses on that pulley? I can't see how this helps me at all. You'll have to forgive me as my physical intutition sucks.
    I don't care much for the answer. I mean, once I have the equations, then the rest is just linear algebra (not that I'm putting down linear algebra or anything).
     
  17. Jun 15, 2004 #16

    arildno

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    A very sensible attitude, which I share!
    (The arithmetic used in solving an n*n linear system isn't exactly the most fascinating subject I know of..)

    What I meant with a detailed solution, was to justify the steps by which I arrive at the set of equations to be used; i.e., deriving them.

    I will derive the set of equations used for a non-slip situation about pulleys with mass, and comment on where simplifications concerning massless pulleys enter.
    The cords/strings are of fixed length, and massless.

    1. Representation of the total force on a pulley from a taut string in terms of the cord tension:
    This is slightly off-topic, but is a justification of the usual technique of substituting values of the cord tension for the net effects of friction and normal force.

    Let the cord tension at a given angle [tex]\theta[/tex] be given in vectorial form as:
    [tex]\vec{T}(\theta)=T(\theta)\vec{i}_{\theta}[/tex]

    That is, at any given point, the tensile force acting on the cord segment ending at [tex]\theta[/tex] is in the tangential direction of increasing angle value.
    Clearly, by Newton's 3.law, the cord segment beginning at [tex]\theta[/tex] will experience a tensile force [tex]-\vec{T}(\theta)[/tex]

    Let us look at a cord segment between
    [tex]\{\theta,\theta+\bigtriangleup\theta\}[/tex]

    Clearly, we may write the total tensile force upon this segment as:
    [tex]\vec{T}(\theta+\bigtriangleup\theta)-\vec{T}(\theta)\approx\frac{d\vec{T}}{d\theta}\bigtriangleup\theta[/tex]

    We also have, by the product rule:
    [tex]\frac{d\vec{T}}{d\theta}=\frac{dT}{d\theta}\vec{i}_{\theta}-T(\theta)\vec{i}_{r}[/tex]

    Using now the condition that the cord is massless, we must have the following equalities between the frictional force from the pulley onto the cord, and the normal force from the pulley onto the cord:
    [tex]\bigtriangleup{F}_{fric,p}=-\frac{dT}{d\theta}\bigtriangleup\theta[/tex]
    [tex]\bigtriangleup{F}_{norm,p}=T(\theta)\bigtriangleup\theta[/tex]

    Clearly, by Newton's 3.law, the total force acting on the pulley from the cord segment is therefore given by:
    [tex]\bigtriangleup\vec{F}_{c}=\frac{d\vec{T}}{d\theta}\bigtriangleup\theta[/tex]

    Hence, the net force from the cord onto the pulley is given by:
    [tex]\vec{F}_{c}=\int_{0}^{\pi}\frac{d\vec{T}}{d\theta}d\theta=\vec{T}(\pi)-\vec{T}(0)[/tex]

    Now the vector tangent at 0 is [tex]\vec{j}[/tex] (assuming that the half-circle of the pulley lies above the contact points), and at [tex]\pi[/tex] is [tex]-\vec{j}[/tex]

    Therefore, the net force on the pulley from the cord can be written as:
    [tex]\vec{F}_{c}=-T_{L}\vec{j}-T_{R}\vec{j}[/tex]

    That is, we can think of the forces acting on the pulley as 2 tensile forces, [tex]\vec{T}_{R},\vec{T}_{L}[/tex] which attack on the right-hand side and left-hand side contact points, respectively.

    This is how the conventional set-up (the one you chose at start!) can be justified.

    More derivations in the next post.
     
  18. Jun 15, 2004 #17

    arildno

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    2. Derivation of equations:
    a) Quantities:
    The two pulleys have masses, radii, moments of inertia given by:
    [tex]M_{1},M_{2},R_{1},R_{2},I_{1},I_{2}[/tex]

    The three blocks have masses [tex]m_{1},m_{2},m_{3}[/tex]

    We will assume that pulley 2 and block 1 go about pulley one; pulley 2 having acceleration [tex]a_{1}[/tex] while block 1 has [tex]-a_{1}[/tex]
    That the accelerations here have the same magnitude follows from the fact that their connecting cord has fixed length.

    Block 2 has a relative acceleration to pulley 2 equal to [tex]a_{2}[/tex] while the relative acceleration of block 3 to pulley 2 is [tex]-a_{2}[/tex]

    The cord tensions are given by:
    [tex]T_{1,L},T_{1,R},T_{2,L},T_{2,R}[/tex]
    where the number indicates about which pulley the cord is laid.
    [tex]T_{1,L}[/tex] works on pulley 2, wheras [tex]T_{1,R}[/tex] works on block 1.

    b) Equations of motion about pulley 1:
    For block 1, we have:
    [tex]T_{1,R}-m_{1}g=-m_{1}a_{1} (1)[/tex]

    For pulley 2, we have:
    [tex]T_{1,L}-(T_{2,L}+T_{2,R})-M_{2}g=M_{2}a_{1} (2)[/tex]

    The frictional torque about pulley 1 must satisfy in the no-slip case:
    [tex]R_{1}(T_{1,R}-T_{1,L})=I_{1}\frac{a_{1}}{R_{1}} (3)[/tex]

    Note that in the case of massless pulley 1, we gain from (3) equality in tensions about it.

    Motion about pulley 2 in the next post.
     
  19. Jun 15, 2004 #18

    arildno

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    3. Motion about pulley 2:
    If you are uncertain about accelerated frames of reference, note that the acceleration of block 2 in the inertial frame fulfills:
    [tex]a_{2,abs}=a_{1}+a_{2}[/tex]
    whereas the acceleration of block 3 in the inertial frame is:
    [tex]a_{3,abs}=a_{1}-a_{2}[/tex]

    Hence, we get:
    [tex]T_{2,L}-m_{2}g=m_{2}a_{2,abs} (4)[/tex]
    [tex]T_{2,R}-m_{3}g=m_{3}a_{3,abs} (5)[/tex]

    The frictional torque about pulley 2 must fulfill:
    [tex]R_{2}(T_{2,R}-T_{1,L})=I_{2}\frac{a_{2}}{R_{2}}(6)[/tex]

    Equations (1)-(6) is the system to be solved.
     
  20. Jun 15, 2004 #19
    That is exactly what I needed! My physics book makes no mention of this, so I was left "in the air". It's funny how my book assigns such problems without explaining anything about Newton's laws in non-inertial frames. I feel cheated here.

    I think you meant
    [tex]R_{2}(T_{2,R}-T_{2,L})=I_{2}\frac{a_{2}}{R_{2}}[/tex]

    Thanks alot. I really appreciate it.
     
  21. Jun 15, 2004 #20

    arildno

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    Yes, that's what I meant..
     
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